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Simplifying a Geometric Series with Two Power Terms
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I have derived a geometric series below that I want to simplify but keep making a mess. Can anyone help?
$$s = aq^n-1r^0 + aq^n-2r^1 + dotsb + aq^1r^n-2 + aq^0r^n-1$$
I get the following, but I know it's wrong
$$s = fracaq^n-1 - (ar^n)/q1 - (r/q)$$
sequences-and-series geometric-series
edited Jul 26 at 22:47
Math Lover
12.3k21232
asked Jul 26 at 22:31
user1753106
132
add a comment |Â
up vote
2
down vote
favorite
I have derived a geometric series below that I want to simplify but keep making a mess. Can anyone help?
$$s = aq^n-1r^0 + aq^n-2r^1 + dotsb + aq^1r^n-2 + aq^0r^n-1$$
I get the following, but I know it's wrong
$$s = fracaq^n-1 - (ar^n)/q1 - (r/q)$$
sequences-and-series geometric-series
edited Jul 26 at 22:47
Math Lover
12.3k21232
asked Jul 26 at 22:31
user1753106
132
Very happy with the answer and hints below. I'm left wondering why my approach of trying to find rs/q like shown on en.wikipedia.org/wiki/Geometric_series didn't work but that's another question I guess ...
– user1753106
Jul 26 at 22:54
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have derived a geometric series below that I want to simplify but keep making a mess. Can anyone help?
$$s = aq^n-1r^0 + aq^n-2r^1 + dotsb + aq^1r^n-2 + aq^0r^n-1$$
I get the following, but I know it's wrong
$$s = fracaq^n-1 - (ar^n)/q1 - (r/q)$$
sequences-and-series geometric-series
edited Jul 26 at 22:47
Math Lover
12.3k21232
asked Jul 26 at 22:31
user1753106
132
I have derived a geometric series below that I want to simplify but keep making a mess. Can anyone help?
$$s = aq^n-1r^0 + aq^n-2r^1 + dotsb + aq^1r^n-2 + aq^0r^n-1$$
I get the following, but I know it's wrong
$$s = fracaq^n-1 - (ar^n)/q1 - (r/q)$$
sequences-and-series geometric-series
edited Jul 26 at 22:47
Math Lover
12.3k21232
asked Jul 26 at 22:31
user1753106
132
edited Jul 26 at 22:47
Math Lover
12.3k21232
edited Jul 26 at 22:47
Math Lover
12.3k21232
edited Jul 26 at 22:47
Math Lover
12.3k21232
12.3k21232
asked Jul 26 at 22:31
user1753106
132
asked Jul 26 at 22:31
user1753106
132
asked Jul 26 at 22:31
user1753106
132
132
Very happy with the answer and hints below. I'm left wondering why my approach of trying to find rs/q like shown on en.wikipedia.org/wiki/Geometric_series didn't work but that's another question I guess ...
– user1753106
Jul 26 at 22:54
add a comment |Â
Very happy with the answer and hints below. I'm left wondering why my approach of trying to find rs/q like shown on en.wikipedia.org/wiki/Geometric_series didn't work but that's another question I guess ...
– user1753106
Jul 26 at 22:54
Very happy with the answer and hints below. I'm left wondering why my approach of trying to find rs/q like shown on en.wikipedia.org/wiki/Geometric_series didn't work but that's another question I guess ...
– user1753106
Jul 26 at 22:54
Very happy with the answer and hints below. I'm left wondering why my approach of trying to find rs/q like shown on en.wikipedia.org/wiki/Geometric_series didn't work but that's another question I guess ...
– user1753106
Jul 26 at 22:54
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
Note that
beginalign
S & = aq^n-1r^0 + aq^n-2r^1 + cdots + aq^1r^n-2 + aq^0r^n-1\
& = aq^n-1left(1+fracrq+ cdots +fracr^n-1q^n-1right)\
& = aq^n-1 frac1-fracr^nq^n1-fracrq \
& = afracq^n-r^nq-r.
endalign
answered Jul 26 at 22:36
Math Lover
12.3k21232
add a comment |Â
up vote
0
down vote
HINT
Recall that we have
$$A^n-B^n=(A-B)(A^n-1+A^n-2B+ldots+AB^n-2+B^n-1)$$
and therefore
$$A^n-1+A^n-2B+ldots+AB^n-2+B^n-1=fracA^n-B^nA-B=A^n-1frac1-left(frac B Aright)^n1-frac B A$$
answered Jul 26 at 22:36
gimusi
65k73583
add a comment |Â
up vote
0
down vote
Hint:
Use the high-school formula:
$$a^n-b^n=(a-b)(a^n-1+a^n-2b+dots+ab^n-2+b^n-1). $$
answered Jul 26 at 22:39
Bernard
110k635102
This got me there myself before @Math Lover thanks. Upvoted but I have no reputation here ..
– user1753106
Jul 26 at 22:51
add a comment |Â
up vote
0
down vote
To find the common ratio (and prove that it is common, and hence a geometric series), divide two successive terms:
$$fracaq^n-1-(k+1)r^k+1aq^n-1-kr^k = fracrq.$$
Note that this doesn't depend on $n$, so the ratio is common between all terms. The initial term is $aq^n-1$, and there are $n$ terms, so the sum is,
$$s = aq^n-1 fracleft(fracrqright)^n - 1fracrq - 1 = aqfracqr^n - 1r - q$$
answered Jul 26 at 22:42
Theo Bendit
11.8k1843
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Note that
beginalign
S & = aq^n-1r^0 + aq^n-2r^1 + cdots + aq^1r^n-2 + aq^0r^n-1\
& = aq^n-1left(1+fracrq+ cdots +fracr^n-1q^n-1right)\
& = aq^n-1 frac1-fracr^nq^n1-fracrq \
& = afracq^n-r^nq-r.
endalign
answered Jul 26 at 22:36
Math Lover
12.3k21232
add a comment |Â
up vote
2
down vote
accepted
Note that
beginalign
S & = aq^n-1r^0 + aq^n-2r^1 + cdots + aq^1r^n-2 + aq^0r^n-1\
& = aq^n-1left(1+fracrq+ cdots +fracr^n-1q^n-1right)\
& = aq^n-1 frac1-fracr^nq^n1-fracrq \
& = afracq^n-r^nq-r.
endalign
answered Jul 26 at 22:36
Math Lover
12.3k21232
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Note that
beginalign
S & = aq^n-1r^0 + aq^n-2r^1 + cdots + aq^1r^n-2 + aq^0r^n-1\
& = aq^n-1left(1+fracrq+ cdots +fracr^n-1q^n-1right)\
& = aq^n-1 frac1-fracr^nq^n1-fracrq \
& = afracq^n-r^nq-r.
endalign
answered Jul 26 at 22:36
Math Lover
12.3k21232
Note that
beginalign
S & = aq^n-1r^0 + aq^n-2r^1 + cdots + aq^1r^n-2 + aq^0r^n-1\
& = aq^n-1left(1+fracrq+ cdots +fracr^n-1q^n-1right)\
& = aq^n-1 frac1-fracr^nq^n1-fracrq \
& = afracq^n-r^nq-r.
endalign
answered Jul 26 at 22:36
Math Lover
12.3k21232
edited Jul 26 at 22:43
answered Jul 26 at 22:36
Math Lover
12.3k21232
answered Jul 26 at 22:36
Math Lover
12.3k21232
answered Jul 26 at 22:36
Math Lover
12.3k21232
12.3k21232
add a comment |Â
add a comment |Â
up vote
0
down vote
HINT
Recall that we have
$$A^n-B^n=(A-B)(A^n-1+A^n-2B+ldots+AB^n-2+B^n-1)$$
and therefore
$$A^n-1+A^n-2B+ldots+AB^n-2+B^n-1=fracA^n-B^nA-B=A^n-1frac1-left(frac B Aright)^n1-frac B A$$
answered Jul 26 at 22:36
gimusi
65k73583
add a comment |Â
up vote
0
down vote
HINT
Recall that we have
$$A^n-B^n=(A-B)(A^n-1+A^n-2B+ldots+AB^n-2+B^n-1)$$
and therefore
$$A^n-1+A^n-2B+ldots+AB^n-2+B^n-1=fracA^n-B^nA-B=A^n-1frac1-left(frac B Aright)^n1-frac B A$$
answered Jul 26 at 22:36
gimusi
65k73583
add a comment |Â
up vote
0
down vote
up vote
0
down vote
HINT
Recall that we have
$$A^n-B^n=(A-B)(A^n-1+A^n-2B+ldots+AB^n-2+B^n-1)$$
and therefore
$$A^n-1+A^n-2B+ldots+AB^n-2+B^n-1=fracA^n-B^nA-B=A^n-1frac1-left(frac B Aright)^n1-frac B A$$
answered Jul 26 at 22:36
gimusi
65k73583
HINT
Recall that we have
$$A^n-B^n=(A-B)(A^n-1+A^n-2B+ldots+AB^n-2+B^n-1)$$
and therefore
$$A^n-1+A^n-2B+ldots+AB^n-2+B^n-1=fracA^n-B^nA-B=A^n-1frac1-left(frac B Aright)^n1-frac B A$$
answered Jul 26 at 22:36
gimusi
65k73583
answered Jul 26 at 22:36
gimusi
65k73583
answered Jul 26 at 22:36
gimusi
65k73583
answered Jul 26 at 22:36
gimusi
65k73583
65k73583
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint:
Use the high-school formula:
$$a^n-b^n=(a-b)(a^n-1+a^n-2b+dots+ab^n-2+b^n-1). $$
answered Jul 26 at 22:39
Bernard
110k635102
This got me there myself before @Math Lover thanks. Upvoted but I have no reputation here ..
– user1753106
Jul 26 at 22:51
add a comment |Â
up vote
0
down vote
Hint:
Use the high-school formula:
$$a^n-b^n=(a-b)(a^n-1+a^n-2b+dots+ab^n-2+b^n-1). $$
answered Jul 26 at 22:39
Bernard
110k635102
This got me there myself before @Math Lover thanks. Upvoted but I have no reputation here ..
– user1753106
Jul 26 at 22:51
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
Use the high-school formula:
$$a^n-b^n=(a-b)(a^n-1+a^n-2b+dots+ab^n-2+b^n-1). $$
answered Jul 26 at 22:39
Bernard
110k635102
Hint:
Use the high-school formula:
$$a^n-b^n=(a-b)(a^n-1+a^n-2b+dots+ab^n-2+b^n-1). $$
answered Jul 26 at 22:39
Bernard
110k635102
answered Jul 26 at 22:39
Bernard
110k635102
answered Jul 26 at 22:39
Bernard
110k635102
answered Jul 26 at 22:39
Bernard
110k635102
110k635102
This got me there myself before @Math Lover thanks. Upvoted but I have no reputation here ..
– user1753106
Jul 26 at 22:51
add a comment |Â
This got me there myself before @Math Lover thanks. Upvoted but I have no reputation here ..
– user1753106
Jul 26 at 22:51
This got me there myself before @Math Lover thanks. Upvoted but I have no reputation here ..
– user1753106
Jul 26 at 22:51
This got me there myself before @Math Lover thanks. Upvoted but I have no reputation here ..
– user1753106
Jul 26 at 22:51
add a comment |Â
up vote
0
down vote
To find the common ratio (and prove that it is common, and hence a geometric series), divide two successive terms:
$$fracaq^n-1-(k+1)r^k+1aq^n-1-kr^k = fracrq.$$
Note that this doesn't depend on $n$, so the ratio is common between all terms. The initial term is $aq^n-1$, and there are $n$ terms, so the sum is,
$$s = aq^n-1 fracleft(fracrqright)^n - 1fracrq - 1 = aqfracqr^n - 1r - q$$
answered Jul 26 at 22:42
Theo Bendit
11.8k1843
add a comment |Â
up vote
0
down vote
To find the common ratio (and prove that it is common, and hence a geometric series), divide two successive terms:
$$fracaq^n-1-(k+1)r^k+1aq^n-1-kr^k = fracrq.$$
Note that this doesn't depend on $n$, so the ratio is common between all terms. The initial term is $aq^n-1$, and there are $n$ terms, so the sum is,
$$s = aq^n-1 fracleft(fracrqright)^n - 1fracrq - 1 = aqfracqr^n - 1r - q$$
answered Jul 26 at 22:42
Theo Bendit
11.8k1843
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To find the common ratio (and prove that it is common, and hence a geometric series), divide two successive terms:
$$fracaq^n-1-(k+1)r^k+1aq^n-1-kr^k = fracrq.$$
Note that this doesn't depend on $n$, so the ratio is common between all terms. The initial term is $aq^n-1$, and there are $n$ terms, so the sum is,
$$s = aq^n-1 fracleft(fracrqright)^n - 1fracrq - 1 = aqfracqr^n - 1r - q$$
answered Jul 26 at 22:42
Theo Bendit
11.8k1843
To find the common ratio (and prove that it is common, and hence a geometric series), divide two successive terms:
$$fracaq^n-1-(k+1)r^k+1aq^n-1-kr^k = fracrq.$$
Note that this doesn't depend on $n$, so the ratio is common between all terms. The initial term is $aq^n-1$, and there are $n$ terms, so the sum is,
$$s = aq^n-1 fracleft(fracrqright)^n - 1fracrq - 1 = aqfracqr^n - 1r - q$$
answered Jul 26 at 22:42
Theo Bendit
11.8k1843
answered Jul 26 at 22:42
Theo Bendit
11.8k1843
answered Jul 26 at 22:42
Theo Bendit
11.8k1843
answered Jul 26 at 22:42
Theo Bendit
11.8k1843
11.8k1843
add a comment |Â
add a comment |Â
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Very happy with the answer and hints below. I'm left wondering why my approach of trying to find rs/q like shown on en.wikipedia.org/wiki/Geometric_series didn't work but that's another question I guess ...
– user1753106
Jul 26 at 22:54